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Fundamental Theorem of Algebra. (1 Viewer)

seanieg89

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That isn't proving the FTA. That is proving that if the FTA holds, then any complex poly of degree n has n roots counting multiplicity. The FTA is USED in the initial step of that induction...

(and it is a perfectly formal way of establishing that fact.)
 
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seanieg89

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Typo: 'constant'--->complex in 5.

Try it guys, it's not too difficult.
 

Carrotsticks

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A few questions:

Could you please explain what you mean exactly by 'minimum value on the disc'? Minimum in terms of what?

Also, for Q2 when you define least positive degree... can I assume nonzero?

Question 3 the polynomial is defined in terms of z, so the x is constant? Or is it pertaining to z = x+iy?
 
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seanieg89

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Lols, yeah. well tweak the questions notation to suit your working...as long as your notation is consistent it doesn't matter too much.

Well meaning there exists a z0 in the disc such that f(z)>=f(z0) for all z in the disc. Think of it as an application of the 2d version of the extreme value theorem.

Huh? Well well yeah the coefficient of this monomial will be nonzero if thats what you mean. If the polynomial in question was 1-x^5+69x^7, then the monomial of least positive degree is -x^5.

Gah. Another typo. x->z.
 
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Pfft.. pure maths. Why don't you do something a little more usefulm with your life?
 

Carrotsticks

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Okay have to head off for the night, but here is what I have so far in summarised form. Will try to post the actual maths later on.

1. Just a bit of working out with Triangle Inequality and remembering that |az| = |a||z| and that |a| is a constant (its the modulus). Once this is established, the bullet point is fairly easy to recognise.

2. Let az^k be some arbitrary constant m and we define z such that z^k = -c^k/m, so it follows that c^k = -1/a. There will be 3 cases to consider for the value of k. First is when k is odd, second is if k is even (but not multiple of 4) and last is if k is a multiple of 4, as we will need to define z according to the case.

3. Was just working on that so haven't 100% established the arithmetic yet, but I can imagine expressing Q(cz) in summation notation, then realising that P(z_0) is constant (so we let it be J or something) and then using the fact that c^k = -1/a to simplify the expression such that we acquire the answer.

4. Similarly to Q1 use the triangle inequality, but remembering that it only works with addition, we write 1 + -z^k instead of 1 - z^k directly.

5. Obtain proof by contradiction because from the fact that |R(z)| < 1.
 
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Very nice math asian, Newton would be proud. Now, recite pi to 100 000 decimal places!
 

seanieg89

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Didn't go to many lectures so I'm not sure...I think the tools of complex analysis you have available by the end of the course make it fairly trivial though. (Like Liouville's theorem.) So it probably appears as a tute question.
 

Carrotsticks

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Yeah pretty sure it will be.

It's in the tutorial (forgot what week).
 

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